Question

A 100 gwire is held under a tension of 250 Nwith one end at x = 0and the other at x = 10.0 m. At time t = 0, pulse 1is sent along the wire from the end at x = 10.0 m. At time t = 30.0 ms, pulse 2is sent along the wire from the end at x = 0.At what position xdo the pulses begin to meet?

Short answer

The position x, at which the pulses begin to meet is 2.63 m from right side and 7.37 m from the left side.

Step-by-step solution

The given data

• Mass of the wire, m = 100 g = 0.1 kg
• Tension in the wire, T = 250 N
• Pulse 1 in sent at t = 0 and from $x_1$=10 m
• Pulse 2 in sent at t = 30 ms = 0.03 s and from $x_2$= 0 m

Understanding the concept of wave equation

When two pulses meet, their coordinates will be the same; hence we will get two equations for position x. By solving them, we can findthevalue of x.

Displacement of a body,

$x=v\times t$ ……..(i)

The velocity of a wave,

$v=\sqrt{\frac{T}{\mu }}$ …….(ii)

Calculation of the position at which pulses meet

Initially, let us calculate velocity of pulses

Since, $\mu =\frac{m}{l}$. The velocity of wave using equation (i) can be given as:

$v=\sqrt{\frac{T\times l}{m}}$

As the wire is held between x = 0 m and x = 10 m, length of the wire will be 10 m, and using value of tension we get, the velocity as:

$$\begin{gathered} v=\sqrt{\frac{250\mathrm{N}\times 10\mathrm{m}}{0.1\mathrm{kg}}} \\ =158\mathrm{m}/\mathrm{s}........................\left(1\right) \end{gathered}$$ role="math" localid="1661155290162" $$\begin{gathered} \mathrm{Pulse}2\mathrm{started}\mathrm{from}{x}_2=0\mathrm{m}\mathrm{at}{t}_2=0.03\mathrm{s} \\ \mathrm{Pulse}1\mathrm{started}\mathrm{from}{x}_1=10\mathrm{m}\mathrm{at}{t}_1=0\mathrm{s} \end{gathered}$$

The figure shows the initial position and time of two pulses, and the position at which two pulses are going to meet.

Now, distance traveled by pulse 1 to reach position x where two pulses meet, is given as-

$$\begin{gathered} x=10m-vt \\ t=\frac{10m-x}{v}..........................\left(2\right) \end{gathered}$$

Now, distance traveled by pulse 2 to reach position x where two pulses meet, is given as-

$x=v(t-t_2).............................\left(3\right)$

Using equation 1, 2, 3 and value of $t_2$, the displacement is given as:

$$\begin{gathered} x=v\left(\frac{10\mathrm{m}-\mathrm{v}}{v}-0.03s\right) \\ =(10m-x)-v\times 0.03s \\ =(10m-x)-(158\mathrm{m}/\mathrm{s}\times 0.03\mathrm{s}) \end{gathered}$$

On solving further,

$$\begin{gathered} 2x=10\mathrm{m}-4.74\mathrm{m} \\ x=\frac{5.26\mathrm{m}}{2} \\ =2.63\mathrm{m} \end{gathered}$$

This is position of x from the right side where the two pulses meet.

From left side the position will be,

$$\begin{gathered} x=10-2.63 \\ =7.37\mathrm{m} \end{gathered}$$

Hence, the position from left and right, where the two pulses meet, are 7.37 m and 2.63 m respectively.